, and M A , R B. , and then draw the shear-force and bending-moment diagrams, labeling all critical ordinates. Solution 10.

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Beatrix Tabitha Jacobs
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1 SETIN 0. ethod of Superposition 63 roblem 0.- The propped cantilever beam shown in the figure supports a uniform load of intensity on the left-hand half of the beam. Find the reactions R, R, and, and then draw the shear-force and bending-moment diagrams, labeling all critical ordinates. R R Solution 0.- Select R ropped cantilever beam SHER-FRE N ENING-ENT IGRS EQUIIRIU R R 8 R R REESE STRUTURE N FRE-ISEENT RETINS x R ( 38 x 7 8 ( ) R 3 3 max TIIITY ( ( ) 0 Substitute for ( and ( ) and solve for R : R 7 8 R max 9 3,768 THER RETINS (FR EQUIIRIU) R roblem 0.-3 The figure shows a propped cantilever beam having span length and an overhang of length a. concentrated load acts at the end of the overhang. etermine the reactions R, R, and for this beam. lso, draw the shear-force and bending-moment diagrams, labeling all critical ordinates. R R a

2 6 HTER 0 Statically Indeterminate eams Solution 0.-3 eam with an overhang Select EQUIIRIU THER RETINS (FR EQUIIRIU) R 3a R ( 3a) R ( a) R ( a) SHER-FRE N ENING-ENT IGRS REESE STRUTURE N FRE-IS. EQS. ( a ( 6 a 3a ( ) ( ) 3 3 a TIIITY ( ( ) 0 Substitute for ( and ( ) and solve for : a roblem 0.- Two flat beams and, lying in horizontal planes, cross at right angles and jointly support a vertical load at their midpoints (see figure). efore the load is applied, the beams just touch each other. oth beams are made of the same material and have the same widths. lso, the ends of both beams are simply supported. The lengths of beams and are and, respectively. What should be the ratio t /t of the thicknesses of the beams if all four reactions are to be the same? t t Solution 0.- Two beams supporting a load For all four reactions to be the same, each beam must support one-half of the load. EFETINS TIIITY or 3 I 3 I ()3 8 ()3 8 ENT F INERTI I bt3 I bt3 3 3 t 3 t 3 t t

3 SETIN 0. ethod of Superposition 6 roblem 0.- etermine the fixed-end moments ( and ) and fixed-end forces (R and R ) for a beam of length supporting a triangular load of maximum intensity 0 (see figure). Then draw the shear-force and bending-moment diagrams, labeling all critical ordinates. R 0 R Solution 0.- Fixed-end beam (triangular load) Select and as reduntants. SYETRY R R 0 96 EQUIIRIU R R 0 o SHER-FRE N ENING-ENT IGRS 0 3 x ( ( ( ) x ( ) REESE STRUTURE N FRE-ISEENT RETINS TIIITY ( ) ( ) 0 Substitute for ( ) and ( ) and solve for :

4 66 HTER 0 Statically Indeterminate eams roblem 0.-6 continuous beam with two uneual spans, one of length and one of length, supports a uniform load of intensity (see figure). etermine the reactions R, R, and R for this beam. lso, draw the shear-force and bending-moment diagrams, labeling all critical ordinates. R R R Solution 0.-6 Select R EQUIIRIU R 3 3 R ontinuous beam with two spans R 3 3 R REESE STRUTURE N FRE-ISEENT RETINS TIIITY ( ( ) 0 R THER RETINS (FR EQUIIRIU) R 8 R 3 6 R 33 6 ( SHER-FRE N ENING-ENT IGRS 9 6 ( ( ( ) R

5 SETIN 0. ethod of Superposition 67 roblem 0.-7 eam is fixed at support and rests (at point ) upon the midpoint of beam E (see the first part of the figure). Thus, beam may be represented as a propped cantilever beam with an overhang and a linearly elastic support of stiffness k at point (see the second part of the figure). The distance from to is 0 ft, the distance from to is / ft, and the length of beam E is 0 ft. oth beams have the same flexural rigidity. concentrated load 700 lb acts at the free end of beam. etermine the reactions R, R, and for beam. lso, draw the shear-force and bending-moment diagrams for beam, labeling all critical ordinates. E = 700 lb k R R = 0 ft = ft Solution 0.-7 Select R EQUIIRIU R R eam with spring support R 3 REESE STRUTURE N FRE-IS. EQS. ( ( ) 7 3 R 3 3 THER RETINS (FR EQUIIRIU) R 7 NUERI UES 700 lb 0 ft 0 in. R 00 lb R 800 lb 30,000 lb-in. SHER-FRE N ENING-ENT IGRS x 300 in. 7.7 in. 3 } TIIITY eam E: 7 3 R 3 3 R 3 8 R ( ( ) R k 8 k 3 R 8 7 () ( IN.) 00 30,000 x 700 0,000

6 68 HTER 0 Statically Indeterminate eams roblem 0.-8 The beam shown in the figure has flexural rigidity.0 N m. When the loads are applied to the beam, the support at settles vertically downward through a distance of 6.0 mm. alculate the reaction R at support. 6 kn/m 3 kn 3 m m m 6 mm settlement R Solution 0.-8 verhanging beam with support settlement Select R settlement of support REESE STRUTURE N FRE-IS. EQS. TIIITY ( ( ) ( ) 3 Substitute for (, ( ), and ( ) 3 and solve for R : R ( 7 08 NUERI UES 6.0 kn/m 3.0 kn 6.0 mm.0 m.0 N m SUSTITUTE INT THE EQUTIN FR R ( ) 3 R 7. kn ( ) 3 R 3 3 R roblem 0.-9 beam is cantilevered from a wall at one end and held by a tie rod at the other end (see figure). The beam is an S 6. I-beam section with length 6 ft. The tie rod has a diameter of / inch and length H 3 ft. oth members are made of steel with E psi. uniform load of intensity 00 lb/ft acts along the length of the beam. efore the load is applied, the tie rod just meets the end of the cable. (a) etermine the tensile force T in the tie rod due to the uniform load. (b) raw the shear-force and bending-moment diagrams for the beam, labeling all critical ordinates. = 00 lb/ft S 6. = 6 ft in. tie rod H = 3 ft

7 SETIN 0. ethod of Superposition 69 Solution 0.-9 eam supported by a tie rod Select the force T in the tie rod REESE STRUTURE N FRE-ISEENT RETINS NUERI UES 00 lb/ft 6 ft H 3 ft E psi eam: S 6. I. in. Tie Rod: d 0. in in. Substitute: T 398 lb ( 8 SHER-FRE N ENING-ENT IGRS T H ( ) ( ) 3 T 3 3 TH E R T 80 lb T (lb) R,30 lb-in. 80 T T x TIIITY ( ( ) ( ) 3 (lb-in.) x or 8 T3 TH 3 E 3 T 8 3 IH,30 x 3.9 in. x. in. roblem 0.-0 The figure shows a nonprismatic, propped cantilever beam with flexural rigidity from to and from to. etermine all reactions of the beam due to the uniform load of intensity. (Hint: Use the results of roblems 9.7- and 9.7-.) R R

8 60 HTER 0 Statically Indeterminate eams Solution 0.-0 Select R Nonprismatic beam FRE-ISEENT RETINS REESE STRUTURE From prob. 9.7-: I S I I S I From prob. 9.7-: I 3 7 I 8 I 7 ( 6 I ( ) 3R 3 6 ( ) downward deflection of end due to load R TIIITY ( ( ) 0 or 7 6 3R R 7 8 ( ) upward deflection due to reaction R EQUIIRIU R R 3 8 R 7 8 roblem 0.- beam is fixed at end and supported by beam E at point (see figure). oth beams have the same cross section and are made of the same material. (a) etermine all reactions due to the load. (b) What is the numerically largest bending moment in either beam? R E R RE

9 SETIN 0. ethod of Superposition 6 Solution 0.- eam supported by a beam et R interaction force between beams select R REESE STRUTURE N FRE-IS. EQS. TIIITY ( ) ( ) ( ) 3 Substitute and solve: R 0 7 ( ) ( R R E ( ( ) ( ) R 3 R 3 38 SYETRY N EQUIIRIU R R E R 0 7 R R R E 3 7 (minus means downward) R 3 7 max E : E E: max max 7 roblem 0.- three-span continuous beam with three eual spans supports a uniform load of intensity (see figure). etermine all reactions of this beam and draw the shear-force and bending-moment diagrams, labeling all critical ordinates. R R R R Solution 0.- Three-span continuous beam SEET R N R S REUNNTS. SYETRY N EQUIIRIU R R REESE STRUTURE FRE-ISEENT RETINS ( ) R ( 3 ( 6 TIIITY R R 3 R ( ( ( ) ) 0 R 0 FRE-ISEENT RETINS ( TIIITY ( ( ) 0 THER RETINS From symmetry and euilibrium: R R 0 R R ( ) R 3 6 R 0 R R R

10 6 HTER 0 Statically Indeterminate eams ING, SHER-FRE, N ENING-ENT IGRS R R R R max roblem 0.-3 beam rests on simple supports at points and (see figure). small gap 0. in. exists between the unloaded beam and a support at point, which is midway between the ends of the beam. The beam has total length 80 in. and flexural rigidity lb-in. lot a graph of the bending moment at the midpoint of the beam as a function of the intensity of the uniform load. Hints: egin by determining the intensity 0 of the load that will just close the gap. Then determine the corresponding bending moment ( ) 0. Next, determine the bending moment (in terms of ) for the case where 0. Finally, make a statically indeterminate analysis and determine the moment (in terms of ) for the case where 0. lot (units of lb-in.) versus (units of lb/in.) with varying from 0 to 00 lb/in. R R = 0 in. = 0. in. = 0 in. R

11 SETIN 0. ethod of Superposition 63 Solution 0.-3 eam on a support with a gap 0 load reuired to close the gap magnitude of gap ( ) bending moment when 0 0 TIIITY or R 3 6 ( ( ) R 6 3 SE 0 EQUIIRIU R R R R 0 R R R R R 3 8 R R R NUERI UES 0. in. 0 in lb-in. Units: lb, in. From es. () and (): lbin. ( ) 0 0,000 lb-in. For 0 : 800 (3) For 0 : 300, () SE 0 0 ( ) 0 0 SE 3 0 (statically indeterminate) Select R REESE STRUTURE 0 ( () () GRH F ENING ENT (EQS. 3 N ) 0 at 00 (lb-in.) 300,000 ( ) 0 0,000 00,000 00,000 00,000 00, EQ. (3) EQ. () at 00 (lb/in.) ( ) R ( ( ) R 3 6

12 6 HTER 0 Statically Indeterminate eams roblem 0.- fixed-end beam of length is subjected to a moment 0 acting at the position shown in the figure. (a) etermine all reactions for this beam. (b) raw shear-force and bending-moment diagrams for the special case in which a b /. R a 0 b R Solution 0.- Fixed-end beam ( 0 applied load) Select R and as redundants. 0 R a b R a b EQUIIRIU R R R 0 REESE STRUTURE N FRE-IS. EQS. ( TIIITY ( ) ( ) ( ) 0 3 or R a(a b) () ( ) ( ) ( ) 0 3 or R 0 a () SE EQS. () N (): R 6 0 ab 3 FR EQUIIRIU: R 6 0 ab 3 0 a (3b ) 0 b (3a ) ( ( 0 a 0 ( 0 a (a b) S SE a b R R ( ) ( ) 3 0 ( ) R ( ) R 3 3 R ( ) 3 ( ) ( ) 3 ( ) 3 (u 0 a (u ) R (u ) 3 ( 0 a (a b) ( ) R 3 3 ( ) 3

Chapter 4.1: Shear and Moment Diagram

Chapter 4.1: Shear and Moment Diagram Chapter 5: Stresses in Beams Chapter 6: Classical Methods Beam Types Generally, beams are classified according to how the beam is supported and according to crosssection